Ch.12 Design via State Space

of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Learning Outcome After completing this chapter, the student will be able to • Design a state-feedback controller using po

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12 Design via State Space

System Dynamics and Control 12.01 Design via State Space

HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien

Learning Outcome

After completing this chapter, the student will be able to

• Design a state-feedback controller using pole placement for systems represented in phase-variable form to meet transient response specifications

• Determine if a system is controllable

• Design a state-feedback controller using pole placement for systems not represented in phase-variable form to meet transient response specifications

• Design a state-feedback observer using pole placement for systems represented in observer canonical form

• Determine if a system is observable

• Design a state-feedback observer using pole placement for systems not represented in observer canonical form

• Design steady-state error characteristics for systems represented in state space

§1.Introduction

- Frequency domain methods of design do not allow to specify all

poles in systems of order higher than 2 because they do not

allow for a sufficient number of unknown parameters to place

all of the closed-loop poles uniquely⟹ State-space methods

solve this problem by introducing into the system (1) other

adjustable parameters and (2) the technique for finding these

parameter values, so that we can properly place all poles of the

closed-loop system

- State-space methods do not allow the specification of

closed-loop zero locations, which frequency domain methods do allow

through placement of the lead compensator zero This is a

disadvantage of state-space methods, since the location of the

zero does affect the transient response Also, a state-space

design may prove to be very sensitive to parameter changes

§1.Introduction

- There is a wide range of computational support for state-space methods; many software packages support the matrix algebra required by the design process However, as mentioned before, the advantages of computer support are balanced by the loss

of graphic insight into a design problem that the frequency domain methods yield

- This chapter considers only an introduction to state-space design only to linear systems

§2.Controller Design

- An𝑛th-order feedback control system has an𝑛th-order

closed-loop characteristic equation of the form

1𝑠𝑛+ 𝑎𝑛−1𝑠𝑛−1+ ⋯ + 𝑎1𝑠 + 𝑎0= 0 (12.1)

Since the coefficient of the highest power of𝑠 isunity, there are

𝑛 coefficients whose values determine the system’s closed-loop

pole locations Thus, if we can introduce 𝑛 adjustable

parameters into the system and relate them to the coefficients

in Eq (12.1), all of the poles of the closed-loop system can be

set to any desired location

§2.Controller Design Topology for Pole Placement

- Consider the closed-loop system represented in state space

ሶ𝒙 = 𝑨𝒙 + 𝑩𝑢 (12.2.a)

ሶ𝒙 =𝑨𝒙 + 𝑩𝑢 = 𝑨𝒙 + 𝑩 −𝑲𝒙 + 𝑟 = 𝑨 − 𝑩𝑲 𝑥 + 𝑩𝑟(12.3.a)

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- A plant signal-flow graph in phase-variable (controller canonical) form

ሶ𝒙 = 𝑨𝒙 + 𝑩𝑢 (12.2.a)

ሶ𝒙 = 𝑨𝒙 + 𝑩𝑢 = 𝑨𝒙 + 𝑩 −𝑲𝒙 + 𝑟 = 𝑨 − 𝑩𝑲 𝑥 + 𝑩𝑟 (12.3.a)

§2.Controller Design Pole Placement for Plants in Phase-Variable Form

- To apply pole-placement methodology to plants represented in phase-variable form

• Represent the plant in phase-variable form

• Feed back each phase variable to the input of the plant through a gain,𝑘𝑖

• Find the characteristic equation for the closed-loop system represented in Step 2

• Decide upon all closed-loop pole locations and determine an equivalent characteristic equation

• Equate like coefficients of the characteristic equations from Steps 3 and 4 and solve for𝑘𝑖

- The phase-variable representation of the plant is given by

𝑨 =

0 1 0 ⋮ 0

0 0 1 ⋮ 0

⋮ ⋮ ⋮ ⋱ ⋮

−𝑎0 −𝑎1 −𝑎2 ⋯ −𝑎𝑛−1

, 𝑩 =

0 0

⋮ 1 , 𝑪 = 𝑐1 𝑐2 ⋯ 𝑐𝑛

The characteristic equation of the plant

𝑠𝑛+ 𝑎𝑛−1𝑠𝑛−1+ ⋯ + 𝑎1𝑠 + 𝑎0= 0 (12.5)

Form the closed-loop system by feeding back each state variable to𝑢

𝑢 = −𝑲𝒙, 𝑘𝑖: the phasevariables’ feedback gains (12.6)

The system matrix,𝑨 − 𝑩𝑲, for the closed-loop system

𝑨−𝑩𝑲=

−(𝑎0+𝑘1) −(𝑎1+𝑘2) −(𝑎2+𝑘3) ⋯ −(𝑎𝑛−1+𝑘𝑛)

(12.8)

(12.4)

- The characteristic equation of

• the plant (the open-loop system)

𝑠𝑛+ 𝑎𝑛−1𝑠𝑛−1+ ⋯ + 𝑎1𝑠 + 𝑎0= 0 (12.5)

• the closed-loop system det 𝑠𝑰 − 𝑨 − 𝑩𝑲 =

𝑠𝑛+ 𝑎𝑛−1+ 𝑘𝑛𝑠𝑛−1+ ⋯+ 𝑎1+ 𝑘2𝑠 + 𝑎0+ 𝑘1 = 0(12.9)

⟹ (12.9) can be derived from (12.5) by adding the appropriate 𝑘𝑖

to each coefficient

- The desired characteristic equation for proper pole placement

𝑠𝑛+ 𝑑𝑛−1𝑠𝑛−1+ 𝑑𝑛−2𝑠𝑛−2+ ⋯+ 𝑑2𝑠2+ 𝑑1𝑠 + 𝑑0= 0(12.10)

- Equating Eqs (12.9) and (12.10) to obtain

𝑑𝑖= 𝑎𝑖+ 𝑘𝑖+1, 𝑖 = 0,1,2, … , 𝑛 − 1

𝑘𝑖+1= 𝑑𝑖− 𝑎𝑖 System Dynamics and Control 12.10 Design via State Space

- For systems represented in phase-variable form, the

numerator polynomial is formed from the coefficients of the

output coupling matrix,𝑪

The plan and closed-loop system are both in phase-variable

form and have the same output coupling matrix ⟹ the

numerators of their transfer functions are the same

𝑠 2 +2𝜁𝜔 𝑛 𝑠+𝜔𝑛, 𝜉 = − ln %𝑂𝑆/100

𝜋 2 +𝑙𝑛 2 %𝑂𝑆/100 , 𝑇 𝑠 =𝜁𝜔4

𝑛

- Ex.12.1 Controller Design for Phase-Variable Form

Design the phase-variable feedback gains to yield%𝑂𝑆 = 9.5%

and𝑇𝑠= 0.74𝑠 Solution The second-order system with the desired performances

𝜉 = − 𝑙𝑛(%𝑂𝑆/100)

𝜋2+ 𝑙𝑛2(%𝑂𝑆/100)= −

𝑙𝑛(9.5/100)

𝜋2+ 𝑙𝑛2(9.5/100)= 0.5996

𝜔𝑛= 4

𝜉𝑇𝑠= 4 0.5996 × 0.74= 9.0147

𝐺 𝑠 = 𝜔𝑛

2

𝑠2+ 2𝜉𝜔𝑛𝑠 + 𝜔𝑛= 81.2648

𝑠 + 5.4 − 𝑗7.2 [(𝑠 + 5.4 + 𝑗7.2 ] The system is third-order⟹ select another closed-loop pole

𝐺 𝑠 = 20(𝑠 + 5) 𝑠(𝑠 + 1)(𝑠 + 4)

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The closed-loop system will have a zero at−5, the same as the

open-loop system⟹ select the third closed-loop pole to cancel

the closed-loop zero𝑝3= −5

However, to demonstrate the effect of the third pole and the

design process, including the need for simulation, let us choose

𝑝3=−5.1

The desired characteristic equation

𝑠 + 5.4 − 𝑗7.2 𝑠 + 5.4 + 𝑗7.2 (𝑠 + 5.1) = 0

⟹ 𝑠3+ 15.9𝑠2+ 136.08𝑠 + 413.1 = 0 (12.17)

Draw the signal-flow diagram for the plant

𝐺 𝑠 = 20(𝑠 + 5) 𝑠(𝑠 + 1)(𝑠 + 4)=

1

𝑠3+5𝑠2+4𝑠 + 0× (20𝑠 + 100)

Feed back all state variables to𝑢 The closed-loopsystem’s state equations

ሶ𝒙 =

−𝑘1 −(4 + 𝑘2) −(5 + 𝑘3)

𝒙 +

0 0 1

𝑟, 𝑦 = 100 20 0 𝒙

The closed-loopsystem’s characteristic equation det 𝑠𝑰 − 𝑨 − 𝑩𝑲 = 𝑠3+ 5 + 𝑘3𝑠2+ 4 + 𝑘2𝑠 + 𝑘1= 0(12.16)

The designed closed-loop and desired characteristic equations

𝑠3+ 5 + 𝑘3𝑠2+ 4 + 𝑘2𝑠 + 𝑘1= 0 (12.16)

𝑠3+ 15.9𝑠2+ 136.08𝑠 + 413.1 = 0 (12.17)

Equating the coefficients of Eqs.(12.16) and (12.17)

𝑘1= 413.1, 𝑘2= 132.08, 𝑘3= 10.9

The state-space representation of the closed-loop system

ሶ𝒙 =

−413.1 −136.08 −15.9

𝒙 +

0 0 1

𝑟 (12.19.a)

𝑦 = 100 20 0 𝒙 (12.19.b)

The closed-loop transfer function

𝑇 𝑠 = 20(𝑠 + 5)

𝑠3+ 15.9𝑠2+ 136.08𝑠 + 413.1

The simulation of the closed-loop system, shows 11.5%

overshoot and a settling time of0.8𝑠 A redesign with the third pole canceling the zero at−5 will yield performance equal to the requirements

Since the steady-state response approaches 0.24 instead of unity, there is a large steady-state error Design techniques to reduce this error are discussed in Section 12.8

Run ch12p1 in Appendix B

Learn how to use MATLAB to

• design a controller for phase variables using pole

placement

• solve Ex.12.1

§2.Controller Design Skill-Assessment Ex.12.1

Problem For the plant

𝐺 𝑠 = 100(𝑠 + 10) 𝑠(𝑠 + 3)(𝑠 + 12) represented in the state space in phase-variable form by

ሶ𝒙 = 00 10 01

0 −36 −15

𝒙 +

0 0 1 𝑟

𝑦 = 1000 100 0 𝒙 design the phase-variable feedback gains to yield5%

overshoot and a peak time of0.3𝑠

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𝐺 𝑠 =𝑠(𝑠+3)(𝑠+12)100(𝑠+10)

Solution The desired characteristic equation

𝜉 = − ln %𝑂𝑆/100

𝜋2+ 𝑙𝑛2%𝑂𝑆/100

= − ln 5/100

𝜋2+ 𝑙𝑛25/100 = 0.69

𝜔𝑛= 𝜋

𝑇𝑝 1 − 𝜉2= 𝜋

0.3 1 − 0.692= 14 47𝑟𝑎𝑑/𝑠

⟹ 𝑠2+ 2𝜉𝜔𝑛𝑠 + 𝜔𝑛2= 𝑠2+ 19.97𝑠 + 209.4

Adding the pole at−10 to cancel the zero at −10

yields the desired characteristic equation

𝑠2+ 19.97𝑠 + 209.4 𝑠 + 10 = 0

⟹ 𝑠3+ 29.97𝑠2+ 409.1𝑠 + 2094 = 0

𝑠 3 + 29.97𝑠 2 + 409.1𝑠 + 2094 = 0

The compensated system matrix in phase-variable form

𝑨 − 𝑩𝑲 =

−𝑘1 −(36 + 𝑘2) −(15 + 𝑘3) The characteristic equation for this system

𝑠𝑰 − 𝑨 − 𝑩𝑲 =

𝑠3+ 15 + 𝑘3𝑠2+ 36 + 𝑘2 𝑠 + 𝑘1

Equating coefficients of this equation with the coefficients of the desired characteristic equation yields the gains as

15 + 𝑘3= 29.97

36 + 𝑘2= 409.1

𝑘1= 14.97

⟹𝐾 = [2094 373.1 14.97]

ሶ𝒙 =

0 1 0

0 0 1

0 −36 −15

𝒙 +

0 0 1 𝑟

𝑦 = 1000 100 0 𝒙

A=[0 1 0; 0 0 1; 0 -36 -15];

B=[0;0;1];

poles=[-3+5j,-3-5j,-10];

K=acker(A,B,poles)

TryIt 12.1

Use MATLAB, the Control

following statements to

solve for the phase-variable

poles of the system in

Skill-Assessment Ex.12.1 at

− 3 − 𝑗5; −3 + 𝑗5, and −10

§3.Controllability

The system is controllable if an input to a system can take every state variable from a desired initial state to a desired final state

Controllability by Inspection

When the system matrix is diagonal, as it is for the parallel form, it is apparent whether or not the system is controllable

ሶ𝒙 =

−𝑎1 0 0

0 −𝑎2 0

0 0 −𝑎3

𝒙 +

1 1 1 𝑢

ሶ𝒙 =

−𝑎4 0 0

0 −𝑎5 0

0 0 −𝑎6

𝒙 +

0 1 1 𝑢

⟹ A system with distinct eigenvalues and a diagonal system matrix is controllable if the input couplingmatrix𝐵 does not have any rows that are zero

§3.Controllability

The Controllability Matrix

An𝑛th-order plant whose state equation is

ሶ𝒙 = 𝑨𝒙 + 𝑩𝑢

is completely controllable if the matrix

𝑪𝑴= [𝑩 𝑨𝑩 𝑨𝟐𝑩 ⋯ 𝑨𝒏−𝟏𝑩]

is of rank𝑛, where 𝑪𝑴is called the controllability matrix

§3.Controllability

- Ex.12.2 Controllability via the Controllability Matrix

Given the system, represented by a signal-flow diagram, determine its controllability

Solution The state equation for the system

ሶ𝒙 = −10 −11 00

0 0 −2

𝒙 +

0 1 1 𝑢

There is the zero in the 𝐵 matrix, this configuration leads to uncontrollabilityonly ifthe poles are real and distinct In this case, the system has multiple poles at−1

The controllability matrix𝑪𝑴= 𝑩 𝑨𝑩 𝑨𝟐𝑩 = 01 −11 −21

1 −2 4

𝑪𝑴 = −1 ≠ 0 ⟹ rank 𝑪𝑴 = 3: the system is controllable

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• test a system for controllability

• solve Ex.12.2

§3.Controllability Skill-Assessment Ex.12.2

Problem Determine whether the system

ሶ𝒙 = 𝑨𝒙 + 𝑩𝑢 =

−1 1 2

0 −1 5

0 3 −4

𝒙 +

2 1 1 𝑢

is controllable Solution The controllability matrix

𝑪𝑴= 𝑩 𝑨𝑩 𝑨𝟐𝑩 =

2 1 1

1 4 −9

1 −1 16

𝑪𝑴 = 80 ≠ 0 ⟹ rank 𝑪𝑴 = 3 : the system is controllable

§3.Controllability

ሶ𝒙 = 𝑨𝒙 + 𝑩𝑢 =

−1 1 0

0 −1 0

0 0 −2

𝒙 +

0 1 1 𝑢

A=[-1 1 2; 0 -1 5; 0 3 -4];

B=[2;1;1];

Cm=ctrb(A,B) Rank=rank(Cm)

TryIt 12.2

Use MATLAB, the Control

following statements to solve

Skill-Assessment Ex.12.2

§4.Alternative Approaches to Controller Design

1stmethod: Matching the coefficients of det(𝑠𝐼 − (𝐴 − 𝐵𝐾)) with

the coefficients of the desired characteristic equation

- Ex.12.3 Controller Design by Matching Coefficients

Design state feedback for the plant represented in cascade form to yield𝑂𝑆% = 15%, 𝑇𝑠= 0.5𝑠

Solution The signal-flow diagram for the plant in cascade form

The state equations

ሶ𝒙 = −𝑘−2 1

1 −(𝑘2+ 1)𝒙 + 01𝑟, 𝑦 = 10 1 𝒙 The characteristics equation

𝑠2+ 𝑘2+ 3 𝑠 + 2𝑘2+ 𝑘1+ 2 = 0 (12.32)

𝐺 𝑠 =𝑌(𝑠) 𝑈(𝑠)=

10 (𝑠 + 1)(𝑠 + 2)

The characteristics equation

𝑠2+ 𝑘2+ 3 𝑠 + 2𝑘2+ 𝑘1+ 2 = 0 (12.32)

The desired characteristic equation

𝜉 = − ln %𝑂𝑆/100

𝜋2+ 𝑙𝑛2%𝑂𝑆/100 = −

ln 15/100

𝜋2+ 𝑙𝑛215/100 = 0.5169

𝜔𝑛= 4

𝑇𝑠𝜉=

4 0.5 × 0.5169= 15 4769𝑟𝑎𝑑/𝑠

⟹ 𝑠2+ 2𝜉𝜔𝑛𝑠 + 𝜔𝑛2=𝑠2+ 16𝑠 + 239.5 = 0 (12.33)

Equating the coefficients of Eqs (12.32) and (12.33)

𝑘2+ 3 = 16

2𝑘2+ 𝑘1+ 2 = 239.5

ቋ ⟹𝑘𝑘1= 211.5

2= 13

2ndmethod:Transforming the system to phase variables, designing

the feedback gains, and transforming the designed system back to its original state-variable representation

- Assume a plant not represented in phase-variable form

ሶ𝒛 = 𝑨𝒛 + 𝑩𝑢, 𝑦 = 𝑪𝒛 (12.34) Controllability matrix

𝑪𝑴𝒛= [𝑩 𝑨𝑩 𝑨𝟐𝑩 ⋯ 𝑨𝒏−𝟏𝑩] (12.35)

- Assume that the system can be transformed into the phase-variable (𝒙) representation with the transformation

Substituting this transformation into Eqs (12.34)

ሶ𝒙 = 𝑷−1𝑨𝑷𝒙 + 𝑷−1𝑩𝑢, 𝑦 = 𝑪𝑷𝒙 (12.37)

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𝑪𝑴𝒛= [𝑩 𝑨𝑩 𝑨 𝟐 𝑩 ⋯ 𝑨 𝒏−𝟏 𝑩] (12.35)

ሶ𝒙 = 𝑷−1𝑨𝑷𝒙 + 𝑷−1𝑩𝑢, 𝑦 = 𝑪𝑷𝒙 (12.37)

Controllability matrix

𝑪𝑴𝒙= [𝑷−1𝑩 𝑷−1𝑨𝑷 𝑷−1𝑩 𝑷−1𝑨𝑷2𝑷−1𝑩 ⋯

𝑷−1𝑨𝑷𝑛−1𝑷−1𝑩 ]

= [𝑷−1𝑩 𝑷−1𝑨𝑷 𝑷−1𝑩 𝑷−1𝑨𝑷 𝑷−1𝑨𝑷 𝑷−1𝑩

⋯ 𝑷−1𝑨𝑷 𝑷−1𝑨𝑷 𝑷−1𝑨𝑷 ⋯ 𝑷−1𝑩 ]

= 𝑷−1[𝑩 𝑨𝑩 𝑨𝟐𝑩 ⋯ 𝑨𝒏−𝟏𝑩] (12.38)

Substituting Eq (12.35) into (12.38) and solving for𝑷

𝑷 = 𝑪𝑴𝒛𝑪𝑴𝒙−1 (12.39)

⟹ the transformation matrix, 𝑷, can be found from the two

controllability matrices

- Design the feedback gains,𝑢 = −𝑲𝒙𝒙 + 𝑟

ሶ𝒙 = 𝑷−1𝑨𝑷𝒙 + 𝑷−1𝑩𝑢

= 𝑷−1𝑨𝑷𝒙 − 𝑷−1𝑩𝑲𝒙𝒙 + 𝑷−1𝑩𝑟

= 𝑷−1𝑨𝑷 − 𝑷−1𝑩𝑲𝒙𝒙 + 𝑷−1𝑩𝑟 (12.40.a)

Since this equation is in phase-variable form, the zeros of this closed-loop system are determined from the polynomial formed from the elements of𝑪𝑷

ሶ𝒙 = 𝑨𝒙 + 𝑩𝑢 = 𝑨𝒙 + 𝑩 −𝑲𝒙 + 𝑟 = 𝑨 − 𝑩𝑲 𝑥 + 𝑩𝑟, 𝑦 = 𝑪𝒙 (12.3)

ሶ𝒙 = 𝑷−1𝑨𝑷 − 𝑷−1𝑩𝑲𝒙𝒙 + 𝑷−1𝑩𝑟, 𝑦 = 𝑪𝑷𝒙 (12.40)

- Transform Eqs (12.40) from phase variables back to the

original representation using𝒙 = 𝑷−1𝒛

ሶ𝒛 = 𝑨𝒛 − 𝑩𝑲𝒙𝑷−1𝒛 + 𝑩𝑟 = 𝑨 − 𝑩𝑲𝒙𝑷−1𝒛 + 𝑩𝑟 (12.41.a)

- Comparing Eqs (12.41) with (12.3), to find the state variable

feedback gain,𝑲𝒛, for the original system

𝑲𝒛= 𝑲𝒙𝑷−1 (12.42)

The TF of this closed-loop system is the same as the TF for

Eqs (12.40), since Eqs (12.40) and (12.41) represent the

same system Thus, the zeros of the closed-loop transfer

function are the same as the zeros of the uncompensated

plant, based upon the development in Section 12.2

- Ex.12.4 Controller Design by Transformation

Design a state-variable feedback controller to yield a 20.8%

overshoot and a settling time of4𝑠 for a plant

𝐺 𝑠 = 𝑠 + 4 (𝑠 + 1)(𝑠 + 2)(𝑠 + 5) that is represented in cascade form

Solution

Original system

The state equations

ሶ𝒛 = 𝑨𝒛𝒛 + 𝑩𝒛𝑢 =

−5 1 0

0 −2 1

0 0 −1

𝒛 +

0 0 1 𝑢

𝑦 = 𝑪𝒛𝒛 = −1 1 0 𝒛

ሶ𝒛 = 𝑨𝒛𝒛 + 𝑩𝒛𝑢 =−50 −21 01

𝒛 +0 1

𝑢, 𝑦 = 𝑪𝒛𝒛 = −1 1 0 𝒛

The controllability matrix

Since det(𝑪𝑴𝒛) = −1 ≠ 0, the system is controllable

The characteristic equation

det 𝑠𝑰 − 𝑨 = 𝑠3+ 8𝑠2+ 17𝑠+ 10= 0

Phase-variable representation of the system

Using the coefficients of the above equation to write

ሶ𝒙 = 𝑨𝒙𝒙 + 𝑩𝒙𝑢 =

0 1 0

0 0 1

−10 −17 −8

𝒙 +

0 0 1 𝑢

𝑦 = 𝑪𝒙𝒙 = 4 1 0 𝒙

𝑪𝑴𝒛= 𝑩𝒛𝑨𝒛𝑩𝒛𝑨𝒛𝑩𝒛 =

0 0 1

0 1 −3

1 −1 1

(𝑠+1)(𝑠+2)(𝑠+5) , 𝑪 𝑴𝒛 =00 01 −31

The output equation was written using the coefficients of the numerator of𝐺(𝑠), since the transfer function must be the same for the two representations

The controllability matrix,𝑪𝑴𝒙, for the phase-variable system

𝑪𝑴𝒙= 𝑩𝒙𝑨𝒙𝑩𝒙𝑨𝒙𝑩𝒙 =

0 0 1

0 1 −8

1 −8 47

(12.48)

Calculate the transformation matrix

The transformation matrix between the two systems

𝑷 = 𝑪𝑴𝒛𝑪𝑴𝒙−1=

1 0 0

5 1 0

10 7 1

(12.49)

Trang 7

Design the controllerusing the phase-variable representation

The desired closed-loop system

• 𝑂𝑆 = 20.8%

𝑇𝑠= 4𝑠

• The closed-loop zero will be at 𝑠 = −4 → choose the third

closed-loop pole to cancel the closed-loop zero

• The total characteristic equation of the desired closed-loop

system

𝐷 𝑠 = 𝑠 + 4 𝑠2+ 2𝑠 + 5

= 𝑠3+ 6𝑠2+ 13𝑠 + 20

ቋ → the designed closed-loop system: 𝑠2+ 2𝑠 + 5

The designed closed-loop system

• The state equations for the phase-variable form with state-variable feedback

ሶ𝒙 = (𝑨𝒙− 𝑩𝒙𝑲𝒙)𝒙 =

−(10 + 𝑘1𝑥) −(17 + 𝑘2𝑥) −(8 + 𝑘3𝑥)𝒙

𝑦 = 𝑪𝒙𝒙 = 4 1 0 𝒙

• The characteristic equation det 𝑠𝑰 − 𝑨𝒙− 𝑩𝒙𝑲𝒙

= 𝑠3+ 8 + 𝑘3 𝑥 𝑠2+ 17 + 𝑘2 𝑥 𝑠 + 10 + 𝑘1 𝑥

Find𝑲𝒙

𝑠3+ 6𝑠2+ 13𝑠 + 20 = 0 (12.50)

𝑠3+ 8 + 𝑘3𝑥𝑠2+ 17 + 𝑘2𝑥 𝑠 + 10 + 𝑘1𝑥 = 0 (12.52)

Comparing Eq.(12.50) with (12.52)

𝑲𝒙= 𝑘1 𝑥𝑘2 𝑥𝑘3 𝑥 = [10 − 4 − 2]

Transform the controller back to the original system

𝑲𝒛= 𝑲𝒙𝑷−1= [−20 10 − 2]

§4.Alternative Approaches to Controller Design Verify the design

The state equations for the designed system

ሶ𝒛 = 𝑨𝒛− 𝑩𝒛𝑲𝒛𝒛 + 𝑩𝒛𝑟 =

−5 1 0

0 −2 1

20 −10 1

𝒛 +

0 0 1 𝑟

𝑇 𝑠 = 𝑠 + 4

𝑠3+ 6𝑠2+ 13𝑠 + 20=

1

𝑠2+ 2𝑠 + 5

𝑦 = 𝑪𝒛𝒛 = −1 1 0 𝒛 The closed-loop TF

• design a controller for a plant not represented in

phase-variable form

• see that MATLAB does not require transformation to

phase-variable form

• solve Ex.12.4

§4.Alternative Approaches to Controller Design Skill-Assessment Ex.12.3

Problem Design a linear state-feedback controller to yield20%

overshoot and a settling time of2𝑠 for a plant

𝐺 𝑠 = 𝑠 + 6 (𝑠 + 9)(𝑠 + 8)(𝑠 + 7) Solution First check controllability

𝑪𝑴𝒛 = −1 ≠ 0 ⟹ rank 𝑪𝑴 = 3 : the system is controllable

Now find the desired characteristic equation

𝑂𝑆 = 20%

𝑇𝑠= 2𝑠

⟹ 𝑠2+ 2𝜉𝜔𝑛𝑠 + 𝜔𝑛2= 𝑠2+ 4𝑠 + 19.24 = 0

ቋ →𝜔𝜉 = 0.456

𝑛= 4.386

𝑪𝑴𝒛= 𝑩 𝑨𝑩 𝑨𝟐𝑩 =

0 0 1

0 1 −17

1 −9 81

Trang 8

𝑠 2 + 4𝑠 + 19.24 = 0

To cancel the zero at−6, adding a pole at −6 yields the

resulting desired characteristic equation

𝑠2+ 4𝑠 + 19.24 𝑠 + 6

= 𝑠3+ 10𝑠2+ 43.24𝑠 + 115.45 = 0

𝐺 𝑠 = 𝑠 + 6

(𝑠 + 9)(𝑠 + 8)(𝑠 + 7)

= 𝑠 + 6

𝑠3+ 24𝑠2+ 191𝑠 + 504

We can write the phase-variable representation

𝑨𝑝=

−504 −191 −24

,𝑩𝑝=

0 0 1 ,𝑪𝑝= [6 1 0]

Since

𝑠 3 + 10𝑠 2 + 43.24𝑠 + 115.45 = 0

The compensated system matrix in phase-variable form

𝑨𝑝− 𝑩𝑝𝑲𝑝=

−(504 + 𝑘1) −(191 + 𝑘2) −(24 + 𝑘3) The characteristic equation for this system

𝑠𝑰 − 𝑨𝑝− 𝑩𝑝𝑲𝑝

= 𝑠3+ 24 + 𝑘3𝑠2+ 191 + 𝑘2𝑠 + (504 + 𝑘1) Equating coefficients of this equation with the coefficients of the desired characteristic equation yields the gains

𝑲𝑝= 𝑘1 𝑘2𝑘3

= [−388.55 − 147.76 − 14]

Now develop the transformation matrix to transform

back to the𝑧-system

𝑪𝑴𝒛= 𝑩𝒛𝑨𝒛𝑩𝒛𝑨𝒛𝑩𝒛 =

0 0 1

0 1 −17

1 −9 81

𝑪𝑴𝒑= 𝑩𝒑𝑨𝒑𝑩𝒑𝑨𝒑2𝑩𝒑 =

0 0 1

0 1 −24

1 −24 385

𝑷 = 𝑪𝑴𝒛𝑪𝑴𝒑−1= 17 01 00

56 15 1

𝑲𝒛= 𝑲𝒑𝑪𝑴𝒑−1

= [−388.55 − 147.76 − 14] 17 01 00

56 15 1

= [−40.23 62.24 − 14]

Therefore

Hence,

§5.Observer Design

- Controller design relies upon access to the state variables for feedback through adjustable gains

- Some of the state variables may not be available at all, or it is too costly to measure them or send them to the controller

- If the state variables are not available because of system configuration or cost, it is possible to estimate the states

Estimated states, rather than actual states, are then fed to the controller One scheme is shown in the figure

An observer, sometimes called an estimator, is used

to calculate state variables that are not accessible from the plant Here the observer

is a model of the plant

§5.Observer Design

-Let’s look at the disadvantages of such configuration Assume the plant

ሶ𝒙 = 𝑨𝒙 + 𝑩𝑢, 𝑦 = 𝑪𝒙 (12.57)

and an observer

ሶෝ𝒙 = 𝑨ෝ𝒙 + 𝑩𝑢, ො𝑦 = 𝑪ෝ𝒙 (12.58)

Subtracting Eqs (12.58) from (12.57) to obtain

ሶ𝒙 − ሶෝ𝒙 = 𝑨(𝒙 − ෝ𝒙), 𝑦 − ො𝑦 = 𝑪(𝒙 − ෝ𝒙) (12.59)

Thus, the dynamics of the difference between the actual and

estimated states is unforced, and if the plant is stable, this

difference, due to differences in initial state vectors, approaches zero

However, the speed of convergence between𝒙 and ෝ𝒙 is too

slow, we seek a way to speed up the observer and make its

response time much faster than that of the controlled

closed-loop system, so that, effectively, the controller will receive the

estimated states instantaneously

§5.Observer Design

- Usefeedbackto increase the speed of convergence between the actual and estimated states

The error between the outputs of the plant and the observer is fed back to the derivatives of the observer’s states The system corrects to drive this error to zero

With feedback we can design a desired transient response into the observer that is much quicker than that of the plant

or controlled closed-loop system

Trang 9

§5.Observer Design

- In designing a controller, the controller canonical (phase-variable) form yields an easy solution for the controller gains In designing an observer, the observer canonical form yields the easy solution for the observer gains

- Example a third-order plant

• represented in observer canonical form

• configured as an observer with the addition of feedback

§5.Observer Design

- The design of the observer is separate from the design of the controller

- Similar to the design of the controller vector,𝑲, the design of the observer consists of evaluating the constant vector,𝑳, so that the transient response of the observer is faster than the response of the controlled loop in order to yield a rapidly updated estimate of the state vector

• Find the state equations for the error between the actual state vector and the estimated state vector,𝒙 − ෝ𝒙

• Find the characteristic equation for the error system and evaluate the required𝑳 to meet a rapid transient response for the observer

§5.Observer Design

- The state equations of the observer

ሶෝ𝒙 = 𝑨ෝ𝒙 + 𝑩𝑢 + 𝑳 𝑦 − ො𝑦 , ො𝑦 = 𝑪ෝ𝒙 (12.60)

- The state equations for the plant

ሶ𝒙 = 𝑨𝒙 + 𝑩𝑢, 𝑦 = 𝑪𝒙 (12.61)

- Subtracting Eqs (12.60) from (12.61) to obtain

ሶ𝒙 − ሶෝ𝒙 = 𝑨 𝒙 − ෝ𝒙 − 𝑳(𝑦 − ො𝑦), 𝑦 − ො𝑦 = 𝑪(𝒙 − ෝ𝒙) (12.62)

𝒙 − ෝ𝒙 : the error between the actual state vector and the

estimated state vector

𝑦 − ො𝑦 : the error between the actual output and the estimated

output

- Subtracting the output equation into the state equation to obtain

ሶ𝒙 − ሶෝ𝒙 = (𝑨 − 𝑳𝑪)(𝒙 − ෝ𝒙), 𝑦 − ො𝑦 = 𝑪(𝒙 − ෝ𝒙) (12.63)

§5.Observer Design

ሶ𝒙 − ሶෝ𝒙 = (𝑨 − 𝑳𝑪)(𝒙 − ෝ𝒙), 𝑦 − ො𝑦 = 𝑪(𝒙 − ෝ𝒙) (12.63)

or ሶ𝒆𝒙= (𝑨 − 𝑳𝑪)𝒆𝒙,𝑦 − ො𝑦 = 𝑪(𝒙 − ෝ𝒙) (12.64)

𝒆𝒙: the estimated state error,𝒆𝒙= 𝒙 − ෝ𝒙

- Equation (12.64a) is unforced If the eigenvalues are all negative, the estimated state vector error,𝒆𝒙, will decay to zero

The design then consists of solving for the values of𝑳 to yield a desired characteristic equation or response for Eqs (12.64)

The characteristic equation is found from Eqs (12.64) to be det 𝜆𝑰 − 𝑨 − 𝑳𝑪 = 0 (12.65) Now we select the eigenvalues of the observer to yield stability and a desired transient response that is faster than the controlled closed-loop response These eigenvalues determine

a characteristic equation that we set equal to Eq (12.65) to solve for𝑳

§5.Observer Design

- Ex.12.5 Observer Design for Observer Canonical Form

Design an observer for the plant

𝐺 𝑠 = 𝑠 + 4

(𝑠 + 1)(𝑠 + 2)(𝑠 + 5)=

𝑠 + 4

𝑠3+ 8𝑠2+ 17𝑠 + 10 which is represented in observer canonical form The observer

will respond10 times faster than the controlled loop designed in

Ex.12.4

Solution

1.First represent the estimated plant in observer canonical form

§5.Observer Design

1.First represent the estimated plant in observer canonical form

2.Now form the difference between theplant’s actual output, 𝑦, and theobserver’s estimated output, ො𝑦, and add the feedback paths from this difference to the derivative of each state variable

Trang 10

§5.Observer Design

3.Next find the characteristic polynomial The state equations for the

estimated plant

ሶො𝒙 = 𝑨ො𝒙 +𝑩𝑢 = −17 0 1−8 1 0

−10 0 0

ො𝒙 +

0 1 4

𝑢, Ƹ𝑦 = 𝑪ො𝒙 = 1 0 0 ො𝒙

The observer error

ሶ𝒆𝑥= 𝑨 − 𝑳𝑪 𝒆𝑥=

−(8 + 𝑙1) 1 0

−(17 + 𝑙2) 0 1

−(10 + 𝑙3) 0 0

𝒆𝑥

The characteristic polynomial

𝑠3+ 8 + 𝑙1𝑠2+ 17 + 𝑙2𝑠 + 10 + 𝑙3 = 0 (12.74)

§5.Observer Design

4.Now evaluate the desired polynomial, set the coefficients equal to those of Eq (12.74), and solve for the gains,𝑙𝑖 From

Eq (12.50), the closed-loop controlled system has dominant second-order poles at−1 ± 𝑗2 To make our observer 10 times faster, we design the observer poles to be at−10 ± 𝑗20 We select the third pole to be 10 times the real part of the dominant second-order poles, or −100 Hence, the desired characteristic polynomial

𝑠 + 100 𝑠2+ 20𝑠 + 500 =

𝑠3+ 120𝑠2+ 2500𝑠 + 50,000 = 0 (12.75) Equating Eqs (12.74) and (12.75) to obtain

𝑙1= 112, 𝑙2= 2483, 𝑙3= 49,990

§5.Observer Design

A simulation of the observer with an input of𝑟 𝑡 = 100𝑡 is

shown in the figure The initial conditions of the plant were all

zero, and the initial condition of𝑥ො1was0.5

Since the dominant pole of the observer is−10 ± 𝑗20, the

expected settling time should be about0.4𝑠 It is interesting to

note the slower response in the figure, where the observer

gains are disconnected, and the observer is simply a copy of

the plant with a different initial condition

§5.Observer Design

Run ch12p4 in Appendix B Learn how to use MATLAB to

• design an observer using pole placement

• solve Ex.12.5

§5.Observer Design

Skill-Assessment Ex.12.4

Problem Design an observer for the plant

𝐺 𝑠 = 𝑠 + 6

(𝑠 + 9)(𝑠 + 8)(𝑠 + 7) whose estimated plant is represented in state space in

observer canonical form as

ሶෝ𝒙 = 𝑨ෝ𝒙 + 𝑩𝑢 = −191 0 1−24 1 0

−504 0 0

ෝ

𝒙 +

0 1 6 𝑢

ො

𝑦 = 𝑪ෝ𝒙 = 1 0 0 ෝ𝒙

The observer will respond 10 times faster than the

controlled loop designed in Skill-Assessment Ex.12.3

§5.Observer Design

Solution The plant is given by

𝐺 𝑠 = 𝑠 + 6 (𝑠 + 9)(𝑠 + 8)(𝑠 + 7)=

20

𝑠3+ 14𝑠2+ 56𝑠 + 64 The characteristic polynomial for the plant with phase-variable state feedback

𝑠3+ (𝑘3+ 14)𝑠2+ (𝑘2+ 56)𝑠 + (𝑘3+ 64) = 0 The desired characteristic equation

𝑠 + 53.33 𝑠2+ 10.67𝑠 + 106.45 =

𝑠3+ 64𝑠2+ 675.48𝑠 + 5676.98 = 0 based upon 15% overshoot, 𝑇𝑠= 0.75𝑠, and a third pole ten times further from the imaginary axis than the dominant poles

Comparing the two characteristic equations

𝑘1= 5612.98, 𝑘2= 619.48, and 𝑘3= 50

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